Find a cubic polynomial p(x) so that the solution of p(x) ≤ 0 is [-2,1] or [4,∞). Is your polynomial the only one that gives this solution? Explain.
clearly you need roots at -2,1,4
Since p < 0 on the first interval, it must be >0 when x < -2.
So, the leading coefficient must be negative, giving us
p(x) = -a(x+2)(x-1)(x-4)
where a is any positive value.
See the graph for a=1 here:
http://www.wolframalpha.com/input/?i=-1(x%2B2)(x-1)(x-4)
To find a cubic polynomial that satisfies the given conditions, we can proceed as follows:
Step 1: Define the intervals of the polynomial
Based on the given conditions, we need the polynomial to be negative (≤ 0) in the intervals [-2, 1] and [4, ∞). We can denote these intervals as I1 = [-2, 1] and I2 = [4, ∞).
Step 2: Determine the equation of the polynomial
A general cubic polynomial can be written as p(x) = ax^3 + bx^2 + cx + d, where a, b, c, and d are constants.
Step 3: Establish inequalities
To guarantee that the polynomial is negative in the given intervals, we need to satisfy the following inequalities:
For I1: p(x) ≤ 0 for -2 ≤ x ≤ 1
This implies that p(-2) ≤ 0 and p(1) ≤ 0.
For I2: p(x) ≤ 0 for x ≥ 4
This implies that p(x) ≤ 0 for any x greater than or equal to 4.
Step 4: Choose an appropriate polynomial
We can choose a simple polynomial that satisfies the given conditions. Let's consider p(x) = (x - 1)(x - 4)(x + 3).
To check if this polynomial indeed satisfies the inequalities, we need to evaluate p(-2), p(1), and p(x) for x ≥ 4.
p(-2) = (-2 - 1)(-2 - 4)(-2 + 3) = (-3)(-6)(1) = 18 > 0
p(1) = (1 - 1)(1 - 4)(1 + 3) = (0)(-3)(4) = 0 ≤ 0
p(x) for x ≥ 4 = (x - 1)(x - 4)(x + 3) = (x - 1)(x + 3)(x - 4) ≤ 0 for x ≥ 4
Since p(1) ≤ 0 and p(x) ≤ 0 for x ≥ 4, the polynomial p(x) = (x - 1)(x - 4)(x + 3) satisfies the given conditions.
Explanation of uniqueness:
Is this polynomial the only one that gives this solution?
No, this polynomial is not the only one that gives this solution. There are infinitely many cubic polynomials that can satisfy the given conditions. We chose one example to demonstrate how to find such a polynomial, but it is not unique.
To find other possible cubic polynomials, you can vary the coefficients or use different factors. Each polynomial with different coefficients or different factors can give rise to a distinct solution that satisfies p(x) ≤ 0 in the specified intervals.